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IGCSE Additional Math2.5. Solve quadratic inequalitiesTopic Practice

2.5. Solve quadratic inequalities

CAIE IGCSE Additional Math 2.5. Solve quadratic inequalities question practice helps you revise this syllabus point with the course map in view. Use this page to focus on one topic, check the style of questions available, and connect each attempt back to the knowledge area it is testing.

EduNinja keeps Additional Math practice aligned to CAIE, so you can move from topic review into exam-style question bank work without losing the syllabus structure. Start with a small set, mark the weak steps, then return to nearby topic links when a definition, graph, calculation, or explanation needs repair.

Question 1

[Maximum number: 3]

Solve the inequality (x+5)(x-2)>3 x+6.

Question 1(b)

[Maximum number: 4]

ALGEBRA

Quadratic Equation

For the equation ax2+bx+c=0a x^{2}+b x+c=0,

x=−b±b2−4ac2ax=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}

Binomial Theorem

(a+b)n=an+(n1)an−1b+(n2)an−2b2+…+(nr)an−rbr+…+bn(a+b)^{n}=a^{n}+\binom{n}{1} a^{n-1} b+\binom{n}{2} a^{n-2} b^{2}+\ldots+\binom{n}{r} a^{n-r} b^{r}+\ldots+b^{n}

where n is a positive integer and (nr)=n!(n−r)!r!\binom{n}{r}=\frac{n!}{(n-r)!r!}

Arithmetic series

un=a+(n−1)dSn=12n(a+l)=12n{2a+(n−1)d}\begin{aligned} & u_{n}=a+(n-1) d \\ & S_{n}=\frac{1}{2} n(a+l)=\frac{1}{2} n\{2 a+(n-1) d\} \end{aligned}

Geometric series

un=arn−1Sn=a(1−rn)1−r(r≠1)S∞=a1−r(∣r∣<1)\begin{aligned} & u_{n}=a r^{n-1} \\ & S_{n}=\frac{a\left(1-r^{n}\right)}{1-r} \quad(r \neq 1) \\ & S_{\infty}=\frac{a}{1-r} \quad(|r|<1) \end{aligned}

2. TRIGONOMETRY

Identities

sin⁡2A+cos⁡2A=1sec⁡2A=1+tan⁡2Acosec⁡2A=1+cot⁡2A\begin{gathered} \sin ^{2} A+\cos ^{2} A=1 \\ \sec ^{2} A=1+\tan ^{2} A \\ \operatorname{cosec}^{2} A=1+\cot ^{2} A \end{gathered}

Formulae for △ABC\triangle A B C

asin⁡A=bsin⁡B=csin⁡Ca2=b2+c2−2bccos⁡AΔ=12bcsin⁡A\begin{gathered} \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C} \\ a^{2}=b^{2}+c^{2}-2 b c \cos A \\ \Delta=\frac{1}{2} b c \sin A \end{gathered}

Solve the inequality 16x−5x2−3<57−9x616 x-5 x^{2}-3<\frac{57-9 x}{6}.

Question 1

[Maximum number: 2]

Solve the inequality (x+2)(4x−5)⩽0(x+2)(4 x-5) \leqslant 0.

Question 1(a)

[Maximum number: 3]

Solve the following inequalities.

x2−x−6⩾0x^{2}-x-6 \geqslant 0

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